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Transcript
```Decidability and Partial
Decidability
• An alphabet is a finite set of symbols.
– Example: a,b,c are symbols; {a,b,c} is an
alphabet.
– Example: 0,1 are symbols; {0,1} is an alphabet.
•  (Sigma) represents an alphabet.
• A string over  is a sequence of symbols
from .
– Example: 01001 is a string over {0,1}
– Example: aabbaaa is a string over {a,b,c}
• * is the set of all finite strings over 
– Example: {0,1}* is the set of all finite binary
strings
• Note that the input to a Turing machine is a
string of tape symbols.
• The output to a Turing machine is also a
string of symbols.
• A problem is a set of yes-no questions.
• A problem consists of a set of objects
(instances of the problem) called the base
set, together with a right answer, either
“yes” or “no”, for each object.
– Example: Blank tape halting problem.
• Base set: set of all Turing machines
• Instance of the problem: A Turing machine M
• Right answer for M: “Yes” if TM M halts on blank
tape, “no” if not
– Example: Student passing problem
• Base set: set of all comp006 students
• Instance of problem: A particular student A
• Right answer for A: “Yes” if student A passes, “no”
if not
• If P is a problem then P, the complement
of P, is P with the “yes” and “no” answers
interchanged.
– The complement of the blank tape halting
problem is the blank tape non-halting problem,
in which the answer is “yes” if the TM fails to
halt on blank tape, and “no” if it halts.
– The complement of the passing problem is the
failing problem.
• A problem may have one or more encodings
which represent instances of the problem as
strings of symbols.
– For the blank tape halting problem an encoding
represents Turing machines as strings of
symbols.
– For the passing problem an encoding represents
students as strings of symbols.
• Let S be the base set of a problem P and let
x be an element of S. A Turing machine M
solves the instance x of P if M gives the
right answer on the encoding x’ of x and
halts.
– M can print “yes” or “no” on the tape
– M can halt in state “y” or “n”
• M solves a problem P if M gives the right
answer on all instances of P, and halts.
– M solves the blank tape halting problem if M,
given an encoding of Turing machine T, always
halts and answers “yes” if T halts on blank tape,
and “no” if not.
– M solves the passing problem if M, given an
encoding of a student A, halts and answers
“yes” if A passes, and halts and answers “no” if
A fails.
• A problem P is decidable (solvable) if there
is a Turing machine T that solves P; such a
T always halts. Else P is undecidable
(unsolvable).
• A problem P is semidecidable (partially
decidable, partially solvable) if there is a
Turing machine T that partially solves P;
such a T solves all instances of P for which
the right answer is “yes”, but fails to halt for
all instances of P for which the right answer
is “no”.
– The blank tape halting problem is
semidecidable if there is a Turing machine M
that, given an encoding of TM T, halts and says
“yes” if T halts on blank tape, but M fails to
halt if T fails to halt on blank tape
– The passing problem is semidecidable if there
is a Turing machine M that, given an encoding
of a student A, halts and says “yes” if A passes
the course, but fails to halt if A fails the course.
– Any problem whose base set is finite, is
decidable! Can you see why? Even if we don’t
know the answer the problem is decidable.
• If P is decidable then P is semidecidable.
• If P is semidecidable and  P is
semidecidable then P is decidable.
• Let Pyes be the set of encodings of instances
x of P such that the right answer is “yes.”
• Let Pno be the set of encodings of instances
x of P such that the right answer is “no.”
• For the blank tape halting problem, Pyes is
the set of encodings of Turing machines that
halt on blank tape. Pno is the set of
encodings of Turing machines that do not
halt on blank tape.
• For the passing problem Pyes is the set of
encodings of students who pass the course
and Pno is the set of encodings of students
who do not pass the course.
• If P is semidecidable then Pyes is called
recursively enumerable.
• If P is decidable, Pyes and Pno are said to be
recursive.
• Functions can also be Turing computable
• A function f from strings to strings is Turing
computable if whenever the Turing machine
starts with x on the tape, it ends with f(x) on
the tape
• Encode integers and tuples of integers as
strings
• Thus one defines computability of functions
on integers
multiplication, subtraction) on integers are
all computable by Turing machines
• Not all functions are computable by Turing
machines.
– The “busy beaver function” is not computable
– We may discuss this later
• All primitive recursive functions are Turing
computable
– We may discuss primitive recursive functions
later
• Some computable functions are not
primitive recursive (Ackermann’s function)
– Maybe we will discuss this later
```